Question

The ac generator in the Figure supplies 150 $V_{\left(max\right)}$ at 50 Hz. With the switch open as shown in the diagram, the resulting current leads the generator emf by ${60}^{\circ }$. With the switch in position 1, the current lags the generator emf by ${30}^{\circ }$. When the switch is in position 2, the maximum current is 3A. Then, the value of R is:

• A. 50/$\sqrt{3}\Omega$
• B. 83.3 $\Omega$
• C. 133.3$\Omega$
• D. 50 $\Omega$

Answer 1

Answer: (A)

50/$\sqrt{3}\Omega$

Answer : A
$switchinposition2:I_{max}=\frac{150}{|L\omega -\frac{1}{C\omega }|}=3\Rightarrow |L\omega -\frac{1}{C\omega }|=50openswitchcondition:\hat{Z}=R+j(L\omega -\frac{1}{C\omega })=Z\angle -60\Rightarrow \frac{R^2}{R^2+{|L\omega -\frac{1}{C\omega }|}^2}={cos}^2(-60)\Rightarrow \frac{R^2}{R^2+{50}^2}={\left(\frac{1}{2}\right)}^2\Rightarrow R=\frac{50}{\sqrt{3}}A$